Integrand size = 24, antiderivative size = 88 \[ \int (c+d x)^2 \csc ^2(a+b x) \sec ^2(a+b x) \, dx=-\frac {2 i (c+d x)^2}{b}-\frac {2 (c+d x)^2 \cot (2 a+2 b x)}{b}+\frac {2 d (c+d x) \log \left (1-e^{4 i (a+b x)}\right )}{b^2}-\frac {i d^2 \operatorname {PolyLog}\left (2,e^{4 i (a+b x)}\right )}{2 b^3} \]
-2*I*(d*x+c)^2/b-2*(d*x+c)^2*cot(2*b*x+2*a)/b+2*d*(d*x+c)*ln(1-exp(4*I*(b* x+a)))/b^2-1/2*I*d^2*polylog(2,exp(4*I*(b*x+a)))/b^3
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(277\) vs. \(2(88)=176\).
Time = 1.88 (sec) , antiderivative size = 277, normalized size of antiderivative = 3.15 \[ \int (c+d x)^2 \csc ^2(a+b x) \sec ^2(a+b x) \, dx=\frac {-\frac {i e^{4 i a} \left (4 b^2 e^{-4 i a} (c+d x)^2+2 i b d \left (1-e^{-4 i a}\right ) (c+d x) \log \left (1-e^{-i (a+b x)}\right )+2 i b d \left (1-e^{-4 i a}\right ) (c+d x) \log \left (1+e^{-i (a+b x)}\right )+2 i b d \left (1-e^{-4 i a}\right ) (c+d x) \log \left (1+e^{-2 i (a+b x)}\right )-2 d^2 \left (1-e^{-4 i a}\right ) \operatorname {PolyLog}\left (2,-e^{-i (a+b x)}\right )-2 d^2 \left (1-e^{-4 i a}\right ) \operatorname {PolyLog}\left (2,e^{-i (a+b x)}\right )-d^2 \left (1-e^{-4 i a}\right ) \operatorname {PolyLog}\left (2,-e^{-2 i (a+b x)}\right )\right )}{-1+e^{4 i a}}+2 b^2 (c+d x)^2 \csc (2 a) \csc (2 (a+b x)) \sin (2 b x)}{b^3} \]
(((-I)*E^((4*I)*a)*((4*b^2*(c + d*x)^2)/E^((4*I)*a) + (2*I)*b*d*(1 - E^((- 4*I)*a))*(c + d*x)*Log[1 - E^((-I)*(a + b*x))] + (2*I)*b*d*(1 - E^((-4*I)* a))*(c + d*x)*Log[1 + E^((-I)*(a + b*x))] + (2*I)*b*d*(1 - E^((-4*I)*a))*( c + d*x)*Log[1 + E^((-2*I)*(a + b*x))] - 2*d^2*(1 - E^((-4*I)*a))*PolyLog[ 2, -E^((-I)*(a + b*x))] - 2*d^2*(1 - E^((-4*I)*a))*PolyLog[2, E^((-I)*(a + b*x))] - d^2*(1 - E^((-4*I)*a))*PolyLog[2, -E^((-2*I)*(a + b*x))]))/(-1 + E^((4*I)*a)) + 2*b^2*(c + d*x)^2*Csc[2*a]*Csc[2*(a + b*x)]*Sin[2*b*x])/b^ 3
Time = 0.56 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.28, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4919, 3042, 4672, 3042, 25, 4202, 2620, 2715, 2838}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int (c+d x)^2 \csc ^2(a+b x) \sec ^2(a+b x) \, dx\) |
| \(\Big \downarrow \) 4919 |
| \(\displaystyle 4 \int (c+d x)^2 \csc ^2(2 a+2 b x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 4 \int (c+d x)^2 \csc (2 a+2 b x)^2dx\) |
| \(\Big \downarrow \) 4672 |
| \(\displaystyle 4 \left (\frac {d \int (c+d x) \cot (2 a+2 b x)dx}{b}-\frac {(c+d x)^2 \cot (2 a+2 b x)}{2 b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 4 \left (\frac {d \int -\left ((c+d x) \tan \left (2 a+2 b x+\frac {\pi }{2}\right )\right )dx}{b}-\frac {(c+d x)^2 \cot (2 a+2 b x)}{2 b}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle 4 \left (-\frac {d \int (c+d x) \tan \left (\frac {1}{2} (4 a+\pi )+2 b x\right )dx}{b}-\frac {(c+d x)^2 \cot (2 a+2 b x)}{2 b}\right )\) |
| \(\Big \downarrow \) 4202 |
| \(\displaystyle 4 \left (-\frac {(c+d x)^2 \cot (2 a+2 b x)}{2 b}-\frac {d \left (\frac {i (c+d x)^2}{2 d}-2 i \int \frac {e^{i (4 a+4 b x+\pi )} (c+d x)}{1+e^{i (4 a+4 b x+\pi )}}dx\right )}{b}\right )\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle 4 \left (-\frac {(c+d x)^2 \cot (2 a+2 b x)}{2 b}-\frac {d \left (\frac {i (c+d x)^2}{2 d}-2 i \left (\frac {i d \int \log \left (1+e^{i (4 a+4 b x+\pi )}\right )dx}{4 b}-\frac {i (c+d x) \log \left (1+e^{i (4 a+4 b x+\pi )}\right )}{4 b}\right )\right )}{b}\right )\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle 4 \left (-\frac {(c+d x)^2 \cot (2 a+2 b x)}{2 b}-\frac {d \left (\frac {i (c+d x)^2}{2 d}-2 i \left (\frac {d \int e^{-i (4 a+4 b x+\pi )} \log \left (1+e^{i (4 a+4 b x+\pi )}\right )de^{i (4 a+4 b x+\pi )}}{16 b^2}-\frac {i (c+d x) \log \left (1+e^{i (4 a+4 b x+\pi )}\right )}{4 b}\right )\right )}{b}\right )\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle 4 \left (-\frac {(c+d x)^2 \cot (2 a+2 b x)}{2 b}-\frac {d \left (\frac {i (c+d x)^2}{2 d}-2 i \left (-\frac {d \operatorname {PolyLog}\left (2,-e^{i (4 a+4 b x+\pi )}\right )}{16 b^2}-\frac {i (c+d x) \log \left (1+e^{i (4 a+4 b x+\pi )}\right )}{4 b}\right )\right )}{b}\right )\) |
4*(-1/2*((c + d*x)^2*Cot[2*a + 2*b*x])/b - (d*(((I/2)*(c + d*x)^2)/d - (2* I)*(((-1/4*I)*(c + d*x)*Log[1 + E^(I*(4*a + Pi + 4*b*x))])/b - (d*PolyLog[ 2, -E^(I*(4*a + Pi + 4*b*x))])/(16*b^2))))/b)
3.3.74.3.1 Defintions of rubi rules used
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[1/(d*e*n*Log[F]) Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) ))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I *((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I Int[(c + d*x)^m*(E^(2*I*( e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt Q[m, 0]
Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp [(-(c + d*x)^m)*(Cot[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1) *Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b _.)*(x_)]^(n_.), x_Symbol] :> Simp[2^n Int[(c + d*x)^m*Csc[2*a + 2*b*x]^n , x], x] /; FreeQ[{a, b, c, d, m}, x] && IntegerQ[n] && RationalQ[m]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (80 ) = 160\).
Time = 1.90 (sec) , antiderivative size = 351, normalized size of antiderivative = 3.99
| method | result | size |
| risch | \(-\frac {2 i d^{2} \operatorname {polylog}\left (2, {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{3}}-\frac {8 d c \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}+\frac {2 d c \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{b^{2}}+\frac {2 d c \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right )}{b^{2}}+\frac {2 d c \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}{b^{2}}-\frac {4 i \left (x^{2} d^{2}+2 c d x +c^{2}\right )}{b \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right ) \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )}-\frac {4 i d^{2} a^{2}}{b^{3}}-\frac {8 i d^{2} x a}{b^{2}}+\frac {2 d^{2} \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right ) x}{b^{2}}-\frac {2 i d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (x b +a \right )}\right )}{b^{3}}+\frac {2 d^{2} \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right ) x}{b^{2}}-\frac {i d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (x b +a \right )}\right )}{b^{3}}+\frac {2 d^{2} \ln \left (1-{\mathrm e}^{i \left (x b +a \right )}\right ) x}{b^{2}}+\frac {2 d^{2} \ln \left (1-{\mathrm e}^{i \left (x b +a \right )}\right ) a}{b^{3}}-\frac {4 i d^{2} x^{2}}{b}+\frac {8 d^{2} a \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{3}}-\frac {2 d^{2} a \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{b^{3}}\) | \(351\) |
-2*I*d^2*polylog(2,exp(I*(b*x+a)))/b^3-8*d/b^2*c*ln(exp(I*(b*x+a)))+2*d/b^ 2*c*ln(exp(I*(b*x+a))-1)+2*d/b^2*c*ln(exp(I*(b*x+a))+1)+2*d/b^2*c*ln(exp(2 *I*(b*x+a))+1)-4*I*(d^2*x^2+2*c*d*x+c^2)/b/(exp(2*I*(b*x+a))+1)/(exp(2*I*( b*x+a))-1)-4*I*d^2/b^3*a^2-8*I*d^2/b^2*x*a+2*d^2/b^2*ln(exp(I*(b*x+a))+1)* x-2*I*d^2/b^3*polylog(2,-exp(I*(b*x+a)))+2*d^2/b^2*ln(exp(2*I*(b*x+a))+1)* x-I*d^2*polylog(2,-exp(2*I*(b*x+a)))/b^3+2*d^2/b^2*ln(1-exp(I*(b*x+a)))*x+ 2*d^2/b^3*ln(1-exp(I*(b*x+a)))*a-4*I*d^2/b*x^2+8*d^2/b^3*a*ln(exp(I*(b*x+a )))-2*d^2/b^3*a*ln(exp(I*(b*x+a))-1)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 950 vs. \(2 (77) = 154\).
Time = 0.32 (sec) , antiderivative size = 950, normalized size of antiderivative = 10.80 \[ \int (c+d x)^2 \csc ^2(a+b x) \sec ^2(a+b x) \, dx=\text {Too large to display} \]
(b^2*d^2*x^2 + 2*b^2*c*d*x - I*d^2*cos(b*x + a)*dilog(cos(b*x + a) + I*sin (b*x + a))*sin(b*x + a) + I*d^2*cos(b*x + a)*dilog(cos(b*x + a) - I*sin(b* x + a))*sin(b*x + a) + I*d^2*cos(b*x + a)*dilog(I*cos(b*x + a) + sin(b*x + a))*sin(b*x + a) - I*d^2*cos(b*x + a)*dilog(I*cos(b*x + a) - sin(b*x + a) )*sin(b*x + a) - I*d^2*cos(b*x + a)*dilog(-I*cos(b*x + a) + sin(b*x + a))* sin(b*x + a) + I*d^2*cos(b*x + a)*dilog(-I*cos(b*x + a) - sin(b*x + a))*si n(b*x + a) + I*d^2*cos(b*x + a)*dilog(-cos(b*x + a) + I*sin(b*x + a))*sin( b*x + a) - I*d^2*cos(b*x + a)*dilog(-cos(b*x + a) - I*sin(b*x + a))*sin(b* x + a) + b^2*c^2 + (b*d^2*x + b*c*d)*cos(b*x + a)*log(cos(b*x + a) + I*sin (b*x + a) + 1)*sin(b*x + a) + (b*c*d - a*d^2)*cos(b*x + a)*log(cos(b*x + a ) + I*sin(b*x + a) + I)*sin(b*x + a) + (b*d^2*x + b*c*d)*cos(b*x + a)*log( cos(b*x + a) - I*sin(b*x + a) + 1)*sin(b*x + a) + (b*c*d - a*d^2)*cos(b*x + a)*log(cos(b*x + a) - I*sin(b*x + a) + I)*sin(b*x + a) + (b*d^2*x + a*d^ 2)*cos(b*x + a)*log(I*cos(b*x + a) + sin(b*x + a) + 1)*sin(b*x + a) + (b*d ^2*x + a*d^2)*cos(b*x + a)*log(I*cos(b*x + a) - sin(b*x + a) + 1)*sin(b*x + a) + (b*d^2*x + a*d^2)*cos(b*x + a)*log(-I*cos(b*x + a) + sin(b*x + a) + 1)*sin(b*x + a) + (b*d^2*x + a*d^2)*cos(b*x + a)*log(-I*cos(b*x + a) - si n(b*x + a) + 1)*sin(b*x + a) + (b*c*d - a*d^2)*cos(b*x + a)*log(-1/2*cos(b *x + a) + 1/2*I*sin(b*x + a) + 1/2)*sin(b*x + a) + (b*c*d - a*d^2)*cos(b*x + a)*log(-1/2*cos(b*x + a) - 1/2*I*sin(b*x + a) + 1/2)*sin(b*x + a) + ...
\[ \int (c+d x)^2 \csc ^2(a+b x) \sec ^2(a+b x) \, dx=\int \left (c + d x\right )^{2} \csc ^{2}{\left (a + b x \right )} \sec ^{2}{\left (a + b x \right )}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 772 vs. \(2 (77) = 154\).
Time = 0.41 (sec) , antiderivative size = 772, normalized size of antiderivative = 8.77 \[ \int (c+d x)^2 \csc ^2(a+b x) \sec ^2(a+b x) \, dx=-\frac {4 \, b^{2} c^{2} + 2 \, {\left (b d^{2} x + b c d - {\left (b d^{2} x + b c d\right )} \cos \left (4 \, b x + 4 \, a\right ) + {\left (-i \, b d^{2} x - i \, b c d\right )} \sin \left (4 \, b x + 4 \, a\right )\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 2 \, {\left (b d^{2} x + b c d - {\left (b d^{2} x + b c d\right )} \cos \left (4 \, b x + 4 \, a\right ) + {\left (-i \, b d^{2} x - i \, b c d\right )} \sin \left (4 \, b x + 4 \, a\right )\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) - 2 \, {\left (b c d \cos \left (4 \, b x + 4 \, a\right ) + i \, b c d \sin \left (4 \, b x + 4 \, a\right ) - b c d\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) - 1\right ) + 2 \, {\left (b d^{2} x \cos \left (4 \, b x + 4 \, a\right ) + i \, b d^{2} x \sin \left (4 \, b x + 4 \, a\right ) - b d^{2} x\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) + 4 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x\right )} \cos \left (4 \, b x + 4 \, a\right ) + {\left (d^{2} \cos \left (4 \, b x + 4 \, a\right ) + i \, d^{2} \sin \left (4 \, b x + 4 \, a\right ) - d^{2}\right )} {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 2 \, {\left (d^{2} \cos \left (4 \, b x + 4 \, a\right ) + i \, d^{2} \sin \left (4 \, b x + 4 \, a\right ) - d^{2}\right )} {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) + 2 \, {\left (d^{2} \cos \left (4 \, b x + 4 \, a\right ) + i \, d^{2} \sin \left (4 \, b x + 4 \, a\right ) - d^{2}\right )} {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) - {\left (i \, b d^{2} x + i \, b c d + {\left (-i \, b d^{2} x - i \, b c d\right )} \cos \left (4 \, b x + 4 \, a\right ) + {\left (b d^{2} x + b c d\right )} \sin \left (4 \, b x + 4 \, a\right )\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - {\left (i \, b d^{2} x + i \, b c d + {\left (-i \, b d^{2} x - i \, b c d\right )} \cos \left (4 \, b x + 4 \, a\right ) + {\left (b d^{2} x + b c d\right )} \sin \left (4 \, b x + 4 \, a\right )\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) - {\left (i \, b d^{2} x + i \, b c d + {\left (-i \, b d^{2} x - i \, b c d\right )} \cos \left (4 \, b x + 4 \, a\right ) + {\left (b d^{2} x + b c d\right )} \sin \left (4 \, b x + 4 \, a\right )\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right ) + 4 \, {\left (i \, b^{2} d^{2} x^{2} + 2 i \, b^{2} c d x\right )} \sin \left (4 \, b x + 4 \, a\right )}{-i \, b^{3} \cos \left (4 \, b x + 4 \, a\right ) + b^{3} \sin \left (4 \, b x + 4 \, a\right ) + i \, b^{3}} \]
-(4*b^2*c^2 + 2*(b*d^2*x + b*c*d - (b*d^2*x + b*c*d)*cos(4*b*x + 4*a) + (- I*b*d^2*x - I*b*c*d)*sin(4*b*x + 4*a))*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) + 2*(b*d^2*x + b*c*d - (b*d^2*x + b*c*d)*cos(4*b*x + 4*a) + ( -I*b*d^2*x - I*b*c*d)*sin(4*b*x + 4*a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) - 2*(b*c*d*cos(4*b*x + 4*a) + I*b*c*d*sin(4*b*x + 4*a) - b*c*d)*arct an2(sin(b*x + a), cos(b*x + a) - 1) + 2*(b*d^2*x*cos(4*b*x + 4*a) + I*b*d^ 2*x*sin(4*b*x + 4*a) - b*d^2*x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + 4*(b^2*d^2*x^2 + 2*b^2*c*d*x)*cos(4*b*x + 4*a) + (d^2*cos(4*b*x + 4*a) + I*d^2*sin(4*b*x + 4*a) - d^2)*dilog(-e^(2*I*b*x + 2*I*a)) + 2*(d^2*cos(4*b *x + 4*a) + I*d^2*sin(4*b*x + 4*a) - d^2)*dilog(-e^(I*b*x + I*a)) + 2*(d^2 *cos(4*b*x + 4*a) + I*d^2*sin(4*b*x + 4*a) - d^2)*dilog(e^(I*b*x + I*a)) - (I*b*d^2*x + I*b*c*d + (-I*b*d^2*x - I*b*c*d)*cos(4*b*x + 4*a) + (b*d^2*x + b*c*d)*sin(4*b*x + 4*a))*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) - (I*b*d^2*x + I*b*c*d + (-I*b*d^2*x - I*b*c*d)*co s(4*b*x + 4*a) + (b*d^2*x + b*c*d)*sin(4*b*x + 4*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) - (I*b*d^2*x + I*b*c*d + (-I*b*d^2*x - I*b*c*d)*cos(4*b*x + 4*a) + (b*d^2*x + b*c*d)*sin(4*b*x + 4*a))*log(cos( b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) + 4*(I*b^2*d^2*x^2 + 2*I *b^2*c*d*x)*sin(4*b*x + 4*a))/(-I*b^3*cos(4*b*x + 4*a) + b^3*sin(4*b*x + 4 *a) + I*b^3)
\[ \int (c+d x)^2 \csc ^2(a+b x) \sec ^2(a+b x) \, dx=\int { {\left (d x + c\right )}^{2} \csc \left (b x + a\right )^{2} \sec \left (b x + a\right )^{2} \,d x } \]
Timed out. \[ \int (c+d x)^2 \csc ^2(a+b x) \sec ^2(a+b x) \, dx=\int \frac {{\left (c+d\,x\right )}^2}{{\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^2} \,d x \]